How is the moment at A using the Force 1200N calculated?

[Benjamin_harsh]

Homework Statement

Find the moment of this Force about point A.

Relevant Equations

$M_{A} = - (1200.Sin 30^{0} X 140) + (1200.Cos 30^{0} X 140)$

A force of 1200N acts on a bracket as shown in the below figure. Find the moment of this Force about point A.

Taking moments of components of 1200N force about point A.

∴ $M_{A} = - (1200.Sin 30^{0} * 140) + (1200.Cos 30^{0} * 140)$

$M_{A} = 40.71 X 10^{3} Nm$

I am confused, because Force 1200 is far away from A and also not touching each other. How is moment at A using Force 1200 is calculated? I am also confused with negative signs.

 

[Doc Al]

Your calculation is flawed: For some reason, you used the same moment arm (140 mm) for each component.

I am confused, because Force 1200 is far away from A and also not touching each other.

So?

How is moment at A using Force 1200 is calculated?

You can break the force into components and find the moment of each, but be sure to use the proper moment arm (which is the perpendicular distance to point A).

I am also confused with negative signs.

Generally, counterclockwise torques are positive; clockwise, negative.

 

[Benjamin_harsh]

proper moment arm (which is the perpendicular distance to point A)

show me the proper moment arm in this diagram.

 

[Doc Al]

Here's a hint: The moment arm for the vertical component will be the horizontal distance; the opposite for the horizontal component.

 

[Benjamin_harsh]

$M_{A} = - (1200.Sin 30^{0} * 140) + (1200.Cos 30^{0} * 1200)$

I got it, what about negative sign? How should I know which sign should be used?

 

[Doc Al]

what about negative sign? How should I know which sign should be used?

Generally, counterclockwise torques are positive; clockwise, negative.

 

[Benjamin_harsh]

Bracket is fixed here, Force 1200N is so far from point A. I can't able to guess the torque directions here. Please help me.

 

[Doc Al]

Bracket is fixed here, Force 1200N is so far from point A. I can't able to guess the torque directions here. Please help me.

The torque directions for each component should be easy to determine.

 

[Benjamin_harsh]

See I constructed X and Y components of force 1200N. Now tell me the torque directions.

 

[Doc Al]

See I constructed X and Y components of force 1200N. Now tell me the torque directions.

You tell me! The vertical component (if it were all that was acting) tends to make things turn in which direction?

 

[Benjamin_harsh]

vertical component (if it were all that was acting) tends to make things turn depends on axis of rotation.

 

[Doc Al]

vertical component (if it were all that was acting) tends to make things turn depends on axis of rotation.

You are finding moments about point A. That's where your axis is.

 

[Benjamin_harsh]

You are finding moments about point A. That's where your axis is.

So torque direction $1200.sin{30^0}$ is counter clockwise direction, It has +ve sign, but in solution it has negative sign, this is the exact point which I am struggling.

 

[Doc Al]

So torque direction $1200.sin{30^0}$ is counter clockwise direction, It has +ve sign, but in solution it has negative sign, this is the exact point which I am struggling.

The standard sign convention is counterclockwise is positive. But perhaps your book uses a different convention. (What book are you using?)

 

[Benjamin_harsh]

The standard sign convention is counterclockwise is positive. But perhaps your book uses a different convention. (What book are you using?)

Are you hinting my torque direction $1200.sin{30^0}$ is correct ?

 

[Doc Al]

Are you hinting my torque direction $1200.sin{30^0}$ is correct ?

That's the correct vertical component and you've drawn the correct torque direction (counterclockwise). The corresponding torque should be +.

 

[Benjamin_harsh]

That's the correct vertical component and you've drawn the correct torque direction (counterclockwise). The corresponding torque should be +.

Then overall moment at A will be negative, but my doubt is can a overall moment can be negative?

 

[Doc Al]

Then overall moment at A will be negative, but my doubt is can a overall moment can be negative?

Why not? (The sign just tells you what direction the torque acts.)

 

[Benjamin_harsh]

The sign just tells you what direction the torque acts.

So I got Moment at A is $40.71 X 10^{3} Nm$ in negative direction, so how does the final position of bracket looks like? Please sketch it.

 

[Doc Al]

So I got Moment at A is $40.71 X 10^{3} Nm$ in negative direction, so how does the final position of bracket looks like? Please sketch it.

Show how you got that answer. (Not sure what you mean by "sketch it".)

 

[sojsail]

how does the final position of bracket looks like?

You are asked to find moment of this Force about point A. Don't assume the bracket is going to rotate to a new position. This would be an analysis of the twisting torque on the bracket (perhaps to see how strong the bracket needs to be considering what will be applied to it in use).

Doc Al said in post #4

Here's a hint: The moment arm for the vertical component will be the horizontal distance; the opposite for the horizontal component.

I notice something suspicious about the figure in your first post. Did you make it or was it part of the problem you were given? The reason is it seems to be seriously not to scale. The vertical position of the point that the force is applied at is shown to be 1200 mm. Could it be that should have been 120 mm? Whichever it is, that dimension should be in your Relevant Equation.

 

[Doc Al]

I notice something suspicious about the figure in your first post. Did you make it or was it part of the problem you were given? The reason is it seems to be seriously not to scale.

Agreed.

 

[CWatters]

Benjamin.. There is a procedure you should use for solving this type of problem.

1) You should start by making a declaration.. "I declare clockwise to be positive" or I declare anticlockwise to be positive. The decision is yours, unless it is stated (or implied) in the problem.

2) Then identify all the torques and their direction and give them the appropriate sign when you write your equation.

3) When you solve the equation the answer will be positive or negative (or zero). Now look back at your declaration to determine what that means (clockwise or anti clockwise).

If you make a different choice of declaration to that in the book _all_ the signs in the working will be opposite. However the direction of the answer (clockwise or anti clockwise) will still be the same. It "all comes out in the wash".

The procedure works no matter what you are trying to solve for. Eg it works if you are trying to find the net torque or if you are given the net and are trying to find an unknown torque.

Ideally your textbook should be consistent throughout and somewhere there should be a statement that says... "Unless stated otherwise, this book declares that (counter) clockwise torques are positive". Unfortunately it is possible your book is not consistent.